1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove existance of open interval

  1. Nov 3, 2009 #1
    Suppose that a function f is differentiable at x0 and that f'(x0)>0. Prove that there exists an open interval containing x0 such that if x1 and x2 are any two points in this interval with x1 < x0 < x2 then f(x1) < f(x0) < f(x2). How do I establish an open interval? Do I need the "epsilon delta" definition of a two sided limit? I am studing maths on my own so please help with this easy question.
    I know, that if a function f is differentiable at x0, then f is continuous at x0 and as f'(x0) is positive then f is an increasing function at x=x0.
     
    Last edited: Nov 3, 2009
  2. jcsd
  3. Nov 4, 2009 #2
    We are given that

    [tex]\lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x - x_0} = L > 0[/tex]

    Let L play the role of epsilon in the definition of this limit.
     
  4. Nov 4, 2009 #3
    Re: Prove existence of open interval

    Limit definition: Let f(x)be defined for all x in some open interval containing the number a, with the possible exception that f(x) need not be defined at a. We will write

    [tex]\lim_{x \rightarrow a}f(x)=L[/tex]

    if given any number epsilon > 0 we can find a number delta > 0 such that
    |f(x) - L| < epsilon if 0 < |x - a| < delta.

    If L plays the role of epsilon in the definition above, what plays the role of |f(x) - L|. I just do not get it yet. Thanks.
     
  5. Nov 4, 2009 #4
    Well just compare what I wrote to your definition. The difference quotient in my limit replaces f(x) in your definition. We replace a with x0 but L remains unchanged since I'm simply using the fact that the limit exists and is > 0. Remember the derivative is just the limit of a difference quotient, and the difference quotient is a function, and so we can just apply the epsilon-delta formulation with the difference quotient as our function.
     
  6. Nov 4, 2009 #5
    Re: Prove existence of open interval

    You mean the following: [tex] \left \frac{f(x) - f(x_{0})}{x - x_{0}} - L \right < \epsilon \mbox{ if } 0 < \left x - x_{0} \right < \delta [/tex].
    So if the conclusion of my definition is true its hypothesis implies the existence of an open interval? Usually theorems work the other way round i.e., you establish its hypothesis and the conclusion follows. Where do I go from here and how do I find a value for delta? Thanks again.
     
  7. Nov 4, 2009 #6
    It's not just the conclusion of the definition of the limit that holds, you have to consider the definition as a whole. Yes usually we have to establish a delta that works given some arbitrary epsilon, but we don't have to do that here since the limit exists. Because the limit exists we can CHOOSE epsilon for our purposes and we know that some delta already exists so that |f(x) - L| is less than the epsilon we chose (clearly we can't actually find a formula for delta since we are working rather generally). As I said before, choose epsilon to be L (since L is a positive number), look carefully at the entire definition of the limit we are working with and figure out where the open intervals come in.
     
  8. Nov 5, 2009 #7
    If you mean
    Let the inverse of [tex] f(x_{0}-L) = a [/tex]
    Let the inverse of [tex] f(x_{0}+L) = b [/tex]
    Then we have the open interval (a,b)
    [tex]x_{0}-\delta =x_{1}, x_{0}+\delta =x_{2} \mbox{ and we have }x_{1} < x_{0} < x_{2} [/tex]
    Then we still have to prove [tex] f(x_{1} < f(x_{0}) < f(x_{2}) [/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Prove existance of open interval
Loading...